\newproblem{lay:2_4_18}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.4.18}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Let $X$ be an $m\times n$ data matrix such that $X^TX$ is invertible, and let $M=I_m-X(X^TX)^{-1}X^T$.
	Add a column $\mathbf{x}_0$ to the data to form $W=\begin{pmatrix}X & \mathbf{x}_0\end{pmatrix}$.
	Compute $W^TW$. The (1,1)-entry is $X^TX$. Show that the Schur complement (Exercise Lay 2.4.16) of
	$X^TX$ can be written in the form $\mathbf{x}_0^TM\mathbf{x}_0$. It can be shown that $(\mathbf{x}_0^TM\mathbf{x}_0)^{-1}$ is
	the (2,2)-entry in $(W^TW)^{-1}$. This entry has a useful statistical interpretation under apropriate hypotheses.
}{
  % Solution
	\begin{center}
		$W^TW=\begin{pmatrix}X^T \\ \mathbf{x}_0^T\end{pmatrix}\begin{pmatrix}X & \mathbf{x}_0\end{pmatrix}=
		 \begin{pmatrix}X^TX & X^T\mathbf{x}_0 \\ \mathbf{x}_0^TX & \mathbf{x}_0^T\mathbf{x}_0 \end{pmatrix}$
	\end{center}
	The Schur complement is defined as $S=A_{22}-A_{21}A_{11}^{-1}A_{12}$, that in this particular case is
	\begin{center}
		$S=\mathbf{x}_0^T\mathbf{x}_0-\mathbf{x}_0^TX(X^TX)^{-1}X^T\mathbf{x}_0=\mathbf{x}_0^T(I_m-X(X^TX)^{-1}X^T)\mathbf{x}_0=\mathbf{x}_0^TM\mathbf{x}_0$
	\end{center}
}
\useproblem{lay:2_4_18}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
